Three boys A,B and C are situated at the vertices of an equilateral triangle of side d at t=0. Each of the boys moves with constant speed v. A always moves towards C and C towards A, At what time and where they meet each other?

Explanation

By symmetry they will meet at the centroid of the triangle. Approaching velocity of A and B towards each other is `v+v cos 60^(@)` and they cover distance d when they meet. So that time taken.
`therefore t=(d)/(v+vcos 60^(@))=(d)/(v+(v)/(2))=(2d)/(3v)`